The equilibrium constant $K_p$  of reaction $N{H}_{4}HS\left(s\right)\cong N{H}_{3\left(g\right)}+H_2{S}_{\left(g\right)}$  is expressed as $x\times {10}^y$ . atm²_. If the observed pressure of mixture in 20 atm what is ‘Y’?

$N{H}_{4}HS\left(s\right)⥂N{H}_3\left(g\right)+H_{2}S\left(g\right)$

t = 0

100 Moles of gases at equilibrium = 2x

t = eq

 1-x x x

$\therefore X_{N{H}_3=\frac{x}{2x}=\frac{1}{2}\&P_{N{H}_3}=\frac{20}{2}=10atm}$

&     $\therefore X_{{}_{H_{2}S=\frac{x}{2x}=\frac{1}{2}\&P_{H_{2}S}=\frac{20}{2}=10atm}}$

$\therefore K_P=10\times 10=100=1.0\times {10}^{2}at{m}^2$